3
$\begingroup$

Let $\mathcal{H}$ be a Hilbert space, and $A$ a self-adjoint operator with domain $D_{A} \subseteq \mathcal{H}$. Assume that $\lambda_0 \in \rho(A)$, where $\rho(A)$ is the resolvent set of $A$. For any $z \in \rho(A)$, let $R_{A}(z)=(A - z I)^{-1}$ be the resolvent of $A$.

Choose $\lambda \neq \lambda_0$. Then it is well known that $\lambda \in \rho(A)$ if and only if $(\lambda - \lambda_0)^{-1} \in \rho(R_{A}(\lambda_0))$ (see e.g. Schmudgen, Unbounded Self-adjoint operators on Hilbert Space, Proposition 2.10). So we have (note that by the spectral theorem $\sigma(A)$ is nonempty): \begin{equation} \sigma(R_{A}(\lambda_0)) \backslash \{0\} = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} If $A$ is a bounded operator on $\mathcal{H}$, then $0 \in \rho(R_{A}(\lambda_0))$, so that in this case, being $\sigma(A)$ closed, we have \begin{equation} \sigma(R_{A}(\lambda_0)) = \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \} = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} Now suppose that $A$ is unbounded. In this case $0 \in \sigma(R_{A}(\lambda_0))$. If we could prove that $0$ is not an isolated point of $\sigma(R_{A}(\lambda_0))$ (which is the same to say that $\sigma(A)$ is not bounded), we could conclude also in this case that \begin{equation} \sigma(R_{A}(\lambda_0)) = \text{closure} \left \{ \frac{1}{\mu - \lambda_0} : \mu \in \sigma(A) \right \}. \end{equation} So my question is the following: if $A$ is unbounded, can $\sigma(A)$ be bounded?

PS This question arouse from the answer given by TrialAndError in this post Norm of the Resolvent

$\endgroup$
  • $\begingroup$ What is $R_A(\lambda_0)$? Is it $(A-\lambda_0)^{-1}$? I think you also have a typo in your first paragraph. Should it be $(\lambda-\lambda_0)^{-1}$? $\endgroup$ – Cameron Williams Jul 11 '16 at 15:15
  • $\begingroup$ $R_A(\mu)$ is usually the inverse of $\mu I - A$ $\endgroup$ – supinf Jul 11 '16 at 15:16
  • 1
    $\begingroup$ This question might be of interest to you: math.stackexchange.com/questions/194681/… $\endgroup$ – Cameron Williams Jul 11 '16 at 15:21
  • $\begingroup$ Dear Cameron, the last answer to the post you quoted is actually related to the argument given by TrialAndError in one of his comments math.stackexchange.com/questions/1855151/norm-of-the-resolvent He actually says that for any normal operator $N$ with a bounded spectrum , we have $ N = - \frac{1}{2 \pi i} \int_{C_{R}} \lambda R_{N}( \lambda) d \lambda $, where $C_R=\{ \lambda \in \mathbb{C}: |\lambda| = R \}$, and $R > 0$ is big enough so that $C_R$ encircles the spectrum of $N$. But I can't see how to prove this representation $\endgroup$ – Maurizio Barbato Jul 11 '16 at 16:00
  • $\begingroup$ If we denote by $\mathcal{B(H)}$ the Banach space of bounded operators on $\mathcal{H}$, then $R_{N}: \rho(N) \rightarrow \mathcal{B(H)}$ is an analytic function. So the integral $\oint_{C_R} \lambda R_{N}(\lambda) d \lambda$ is well defined and by definition is in $\mathcal{B(H)}$. So if we could prove the above representation for $N$, then we could conclude that $N$ cannot be unbounded if it has a bounded spectrum. $\endgroup$ – Maurizio Barbato Jul 11 '16 at 16:06
4
$\begingroup$

Take $r > \max \sigma(A)$. Then $R_A(r)$ is self-adjoint and bounded. If $0$ is not in its spectrum, then $A = (R_A(r)^{-1}+rI$ is bounded. If $0$ is in its spectrum, it is an isolated point of the spectrum and therefore must be an eigenvalue: $R_A v = 0$ for some $v \in \mathcal H$. But that is impossible since $R_A(r) = (A-rI)^{-1}$, i.e. $R_A(r) v = u$ where $u \in D_A$ and $(A-rI) u = v$.

$\endgroup$
  • $\begingroup$ Wonderful proof, dear prof. Israel! Thank you vercy much for your help! $\endgroup$ – Maurizio Barbato Jul 11 '16 at 15:49
4
$\begingroup$

You mentioned you like Complex Analysis. So I thought I'd offer a proof using Complex Analysis applied to the resolvent. The proof comes down to evaluating the integral around all finite singularities of $(\lambda I-A)^{-1}x$ by determining the residue at $\infty$, which turns out to be $x$. This equivalence forces the completeness of spectral expansions for normal operators. It's a type of Complex Analysis conservation law that allows you to expand $x$ in terms of integrals in the finite plane. If the singularities are all discrete in the finite plane, you end up with an eigenfunction expansion of $x$. Continuous spectrum can lead to integral expansions, such as the classical Fourier integral expansions. More generally, the Spectral Theorem for sefadjoint operators can be proved using this conversation law; completeness is established by knowing that the resiude at infinity of $(\lambda I-A)^{-1}$ is $I$. So the technique is worth learning.

Suppose $A$ is a closed densely-defined normal operator on a Complex Hilbert space $\mathcal{H}$, and suppose that $\sigma(A)$ is a bounded set. By the previous problem you referenced, $$ \|(\lambda I -A)^{-1}\| \le \frac{1}{\mbox{dist}(\lambda,\sigma(A))}. $$ Therefore, $\lim_{\lambda\rightarrow\infty} (\lambda I-A)^{-1}=0$, and, for a fixed $x\in\mathcal{D}(A)$, the following limit is uniform in $\lambda$: $$ \lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}x=\lim_{\lambda\rightarrow\infty}x+(\lambda I-A)^{-1}Ax = x. $$ If $x\in\mathcal{D}(A)$, and if $R$ is large enough that $\sigma(A)\subseteq \{ \lambda : |\lambda| < R \}$, then $$ \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}xd\lambda = \lim_{R\rightarrow\infty}\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda(\lambda I-A)^{-1}x\frac{d\lambda}{\lambda}=x. $$ Because $\mathcal{D}(A)$ is dense and $\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda$ is a bounded operator, then $$ \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}d\lambda =I. $$ For $x\in\mathcal{D}(A)$, \begin{align} Ax & = \frac{1}{2\pi i}\oint_{|\lambda|=R}(\lambda I-A)^{-1}Ax\,d\lambda \\ & = \frac{1}{2\pi i}\oint_{|\lambda|=R}-x+\lambda(\lambda I-A)^{-1}x\,d\lambda \\ & = \left(\frac{1}{2\pi i}\oint_{|\lambda|=R}\lambda (\lambda I-A)^{-1}d\lambda\right)x \end{align} So $A$ is bounded on $\mathcal{D}(A)$, which also forces $\mathcal{D}(A)=\mathcal{H}$ because $A$ is closed.

$\endgroup$
  • $\begingroup$ It is such a wonderful application of holomorphic calculus that I am still astonished. I don't know how to thank you for having so generously given all the details of this incredibly beautiful proof: a true pearl! $\endgroup$ – Maurizio Barbato Jul 12 '16 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.